Are you familiar with the Taylor series? That's the first organ of the Taylor series, something like two decades ago I checked how accurate it goes past 20 organs:

https://dotancohen.com/eng/taylor-sine.php

Since e^(2πi) = 1, we can also conclude that e^(2πifx) = 1^(fx). This makes Complex Fourier Transforms quite trivial.

pi = 3.2

(that is an assignment statement)

https://en.wikipedia.org/wiki/Indiana_pi_bill

I am not sure I understand. Sin(x) approaches x only when x approaches 0. When else does the universe does a bad job with this approximation?

https://news.ycombinator.com/item?id=42174105

New submission here.

I came here to say the same thing!

YouTube has become a fantastic place for this long tail of content, in this particular case a bunch of interesting math problems and tricks presented on a blackboard. Or, even full classes, from a person focused on honing pedagogy.

3blue1brown is another amazing channel for math as well.

I have a feeling that this sort of content is the seeds of very great things for humanity. In the 20th century, ET Jaynes talks about how people never get credit in academia for creating simpler paths to greater understanding. But with YouTube, creators can both reach an audience and also find patrons to support them, or maybe even make a living off of YouTube directly with enough viewers.

Motivated students have such resources at their fingertips just from an internet connection, if they happen to get lucky enough to find the right resources.

I wonder what the easiest to prove example of a, b irrational with a^b rational is?

The easiest I can think of offhand would be e^log(2). To prove that we need to prove that e is irrational and the log(2) is irrational.

To prove log(2) is irrational one approach is to prove that e^r is irrational for rational r != 0, which would imply that if log(2) is rational then e^log(2) would be irrational. To prove that e^r is irrational for irrational r it suffices to prove that e^n is irrational for all positive integers n.

We'd also get the e is irrational out of that by taking n = 1, and that would complete our proof that e^log(2) is an example of irrational a, b with a^b rational.

So, all we need now is a proof that e^n is irrational for integers n > 0.

The techniques used in Niven's simple proof that pi is irrational, which was discussed here [1], can be generalized to e^n. You can find that proof in Niven's book "Irrational Numbers" or in Aigner & Ziegler's "Proofs from THE BOOK".

That can also be proved by proving that e is transcendental. Normally proofs that specific numbers are transcendental (other than numbers specifically constructed to be transcendental) are fairly advanced but for e you can do it with first year undergraduate calculus. There's a chapter in Spivak's "Calculus" that does it, and there's a proof in the aforementioned "Irrational Numbers".

[1] https://news.ycombinator.com/item?id=41178560

pretty sure you have a typo, should be "If not, consider (sqrt(2) ^ sqrt(2)) ^ sqrt(2)."

Well the proof I would use is let a = e and b = i(pi).

e^(i theta) = cos theta + i sin theta (Euler's identity) thus e^(i pi) = cos pi + i sin pi = -1 + i(0) = -1

We know that e and i pi are irrational (in fact i pi isn't even a real) and -1 is rational.

Therefore there exist two numbers a and b such that both a and b are irrational but a^b is rational.

In fact log of just about anything is irrational so e^(log x) works as well for just about all rational x, but Euler's identity is cool so I wanted to use that.

Small but important correction: no polynomial with integer coefficients (equivalently, rational coefficients). p(x) = (x - 3^(sqrt(3))) is a perfectly fine polynomial with real coefficients.

No polynomial with rational coefficients. Of course x-y evaluates to 0 when x=y, even if y is a transcendental number.

Good idea. I think I've rushed into publishing this. Anyway, I've reposted the link, and I will do the graph you've mentioned.

An unfortunate mistake...